A bag contains 23 balls, of which 7 are identical. The number of ways of selecting 12 balls from the bag is....

Total number of balls = $23$
Number of identical balls = $7$
Number of distinct balls = $23 - 7 = 16$

We need to select $12$ balls. Let the number of identical balls selected be $k$, where $0 \le k \le 7$.
The remaining $12 - k$ balls must be selected from the $16$ distinct balls. The number of ways to do this is ${}^{16}C_{12-k}$.

The total number of ways to select $12$ balls is the sum of these possibilities for $k = 0$ to $7$:
$S = \sum_{k=0}^{7} {}^{16}C_{12-k}$
$S = {}^{16}C_{12} + {}^{16}C_{11} + {}^{16}C_{10} + {}^{16}C_{9} + {}^{16}C_{8} + {}^{16}C_{7} + {}^{16}C_{6} + {}^{16}C_{5}$

Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$, we can rewrite the first four terms:
${}^{16}C_{12} = {}^{16}C_{4}$
${}^{16}C_{11} = {}^{16}C_{5}$
${}^{16}C_{10} = {}^{16}C_{6}$
${}^{16}C_{9} = {}^{16}C_{7}$

Substituting these back into the sum:
$S = {}^{16}C_{4} + {}^{16}C_{5} + {}^{16}C_{6} + {}^{16}C_{7} + {}^{16}C_{8} + {}^{16}C_{7} + {}^{16}C_{6} + {}^{16}C_{5}$

Rearranging and grouping the terms:
$S = ({}^{16}C_{4} + {}^{16}C_{5}) + ({}^{16}C_{5} + {}^{16}C_{6}) + ({}^{16}C_{6} + {}^{16}C_{7}) + ({}^{16}C_{7} + {}^{16}C_{8})$

Using Pascal's identity ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$:
$S = {}^{17}C_{5} + {}^{17}C_{6} + {}^{17}C_{7} + {}^{17}C_{8}$

Grouping again:
$S = ({}^{17}C_{5} + {}^{17}C_{6}) + ({}^{17}C_{7} + {}^{17}C_{8})$
$S = {}^{18}C_{6} + {}^{18}C_{8}$

Answer: ${}^{18}C_6 + {}^{18}C_8$